2015 AMC 10A Problems/Problem 25

Problem 25

Let $S$ be a square of side length $1$ . Two points are chosen independently at random on the sides of $S$ . The probability that the straight-line distance between the points is at least $\dfrac{1}{2}$ is $\dfrac{a-b\pi}{c}$ , where $a$ , $b$ , and $c$ are positive integers with $\gcd(a,b,c)=1$ . What is $a+b+c$ ?

$\textbf{(A) }59\qquad\textbf{(B) }60\qquad\textbf{(C) }61\qquad\textbf{(D) }62\qquad\textbf{(E) }63$

Solution

Divide the boundary of the square into halves, thereby forming $8$ segments. Without loss of generality, let the first point $A$ be in the bottom-left segment. Then, it is easy to see that any point in the $5$ segments not bordering the bottom-left segment will be distance at least $\dfrac{1}{2}$ apart from $A$ . Now, consider choosing the second point on the bottom-right segment. The probability for it to be distance at least $0.5$ apart from $A$ is $\dfrac{0 + 1}{2} = \dfrac{1}{2}$ because of linearity of the given probability. (Alternatively, one can set up a coordinate system and use geometric probability.)

If the second point $B$ is on the left-bottom segment, then if $A$ is distance $x$ away from the left-bottom vertex, then $B$ must be at least $\dfrac{1}{2} - \sqrt{0.25 - x^2}$ away from that same vertex. Thus, using an averaging argument we find that the probability in this case is $\frac{1}{\left( \frac{1}{2} \right)^2} \int_0^{\frac{1}{2}} \dfrac{1}{2} - \sqrt{0.25 - x^2} dx = 4\left( \frac{1}{4} - \frac{\pi}{16} \right) = 1 - \frac{\pi}{4}.$

(Alternatively, one can equate the problem to finding all valid $(x, y)$ with $0 < x, y < \dfrac{1}{2}$ such that $x^2 + y^2 \ge \dfrac{1}{4}$ , i.e. $(x, y)$ is outside the unit circle with radius $0.5.$ )

Thus, averaging the probabilities gives $P = \frac{1}{8} \left( 5 + \frac{1}{2} + 1 - \frac{\pi}{4} \right) = \frac{1}{32} \left( 26 - \pi \right).$

Our answer is $\boxed{\textbf{(A) } 59}$ .

Solution 2

Let one point be chosen on a fixed side. Then the probability that the second point is chosen on the same side is $\frac{1}{4}$ , on an adjacent side is $\frac{1}{2}$ , and on the opposite side is $\frac{1}{4}$ . We discuss these three cases.

Case 1: Two points are on the same side. Let the first point be $a$ and the second point be $b$ in the $x$ -axis with $0\le a, b\le 1$ . Consider $(a, b)$ a point on the unit square $[0,1]\times [0,1]$ on the $(x, y)$ -plane. The region $\{(a,b): |b-a|> \frac{1}{2}\}$ has the area of $(\frac{1}{2})^2$ . Therefore, the probability that $|b-a|> \frac{1}{2}$ is $\frac{1}{4}$ .

Case 2: Two points are on two adjacent sides. Let the two sides be $[0,1]$ on the x-axis and $[0,1]$ on the y-axis and let one point be $(a, 0)$ and the other point be $(0, b)$ . Then $0\le a, b\le 1$ and the distance between the two points is $\sqrt{a^2+b^2}$ . As in Case 1, $(a, b)$ is a point on the unit square $[0,1]\times [0,1]$ . The area of the region $\{(a,b): \sqrt{a^2+b^2} \le 1/2, 0\le a, b\le 1\}$ is $\pi/16$ and the area of its complementary set inside the square (i.e. $\{(a,b): \sqrt{a^2+b^2} > 1/2, 0\le a, b\le 1\}$ ) is $1-\pi/16$ . . Therefore, the probability that the distance between $(a, 0)$ and $(0, b)$ is at least $1/2$ is $1-\pi/16$ .

Case 3: Two points are on two opposite sides. In this case, the probability that the distance between the two points is at least $1/2$ is obviously $1$ .

Thus the probability that the probability that the distance between the two points is at least $1/2$ is given by $\frac{1}{4} \cdot \frac{1}{4}+ \frac{1}{2}(1 - \frac{\pi}{16}) + \frac{1}{4} =\frac{26-\pi}{32}.$ Therefore $a=26$ , $b=1$ , and $c=32$ . Thus, $a+b+c=59$ and the answer is $\text{(A).}$

2015 AMC 10A (Problems • Answer Key • Resources)
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2015 AMC 10A Problems/Problem 25

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Problem 25

Solution

Solution 2

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